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25b that δ (t) ↔ iω and δ (t) ↔ −ω 2 , so −δ (t) + 2iδ (t) + δ(t) ↔ ω 2 − 2ω + 1. Answers to selected exercises for chapter 9 39 c Since (ω − 1)2 = ω 2 − 2ω + 1, the results in part a and b should be the same. 30b with f (t) = eit we indeed obtain that eit δ (t) = δ (t) − 2iδ (t) − δ(t). 25b it follows that δ (t) ↔ −ω 2 . The convolution theorem then implies that T ∗ δ (t) ↔ −ω 2 U where U is the spectrum of T . 21). b As noted in part a we have that δ ∗ | t | = | t | . 15c it was shown that | t | = 2δ, so we indeed get δ ∗ | t | = 2δ.

N=1 Here we have also split a sum in terms with n = 0, n > 1 and n < −1, and then changed from n to −n in the sum with n < −1. The sums in the righthand side are geometric series with ratio r = e−2π(a+it) and r = e−2π(a−it) respectively. Note that | r | < 1 since a > 0. 16), then writing the result with a common denominator, and finally multiplying everything out and simplifying, it follows that ∞ a X 1 1 − e−4πa . = 2 2 −4πa π n=−∞ a + (t + n) 1+e − e−2πa (e2πit + e−2πit ) Multiplying numerator and denominator by e2πa the result follows.

Now if −a ≤ t ≤ a, then t − a ≤ 0 ≤ t + a and so Z t+a Z 0 u −| v | e−u du = 2 − 2e−a cosh t. e du + (p2a (v) ∗ e )(t) = t−a 0 If t > a, then t − a > 0 and so Z t+a e−u du = 2e−t sinh a. (p2a (v) ∗ e−| v | )(t) = t−a Finally, if t < −a, then t + a < 0 and so Z t+a eu du = 2et sinh a. (p2a (v) ∗ e−| v | )(t) = t−a c The function g from part b is continuous at t = a since limt↓a g(t) = 2e−a sinh a = e−a (ea − e−a ) = 1 − e−2a and g(a) = limt↑a g(t) = 2 − 2e−a cosh a = 2 − e−a (ea + e−a ) = 1 − e−2a .

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